Problem: Which of the following numbers is a multiple of 8? ${54,58,91,105,120}$
The multiples of $8$ are $8$ $16$ $24$ $32$ ..... In general, any number that leaves no remainder when divided by $8$ is considered a multiple of $8$ We can start by dividing each of our answer choices by $8$ $54 \div 8 = 6\text{ R }6$ $58 \div 8 = 7\text{ R }2$ $91 \div 8 = 11\text{ R }3$ $105 \div 8 = 13\text{ R }1$ $120 \div 8 = 15$ The only answer choice that leaves no remainder after the division is $120$ $ 15$ $8$ $120$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $8$ are contained within the prime factors of $120$ $120 = 2\times2\times2\times3\times5 8 = 2\times2\times2$ Therefore the only multiple of $8$ out of our choices is $120$. We can say that $120$ is divisible by $8$.